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Math Help - Surface integral problem

  1. #1
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    Surface integral problem

    I'm having a little trouble with finishing a surface integral problem. The problem statement is this:

    F(x,y,z) = ix + yj + zk, where S is the hemisphere z = (a^2 - x^2 - y^2)^(1/2).

    The answer is supposed to be 2pi*a^3, but I'm stuck right after switching to polar coordinates. I've attached a picture of what I've done so far.

    Thanks for any help!

    Surface integral problem-surface_integral.jpg
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  2. #2
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    Re: Surface integral problem

    Just a quick look: $\vec{T}_s$ is wrong. It should be $\vec{T}_s = (1,0,-s(a^2-s^2-t^2)^{-1/2})$ (and similarly for $\vec{T}_t$).

    I took a closer look. It won't affect your calculations. You have:

    $\displaystyle \int\int_D\left[(s^2+t^2)(a^2-s^2-t^2)^{-1/2} + (a^2-s^2-t^2)^{1/2}\right]dsdt$

    I will rewrite this:

    $\dfrac{s^2+t^2}{(a^2-s^2-t^2)^{1/2}} + (a^2-s^2-t^2)^{1/2} = \dfrac{s^2+t^2 + (a^2-s^2-t^2)}{(a^2-s^2-t^2)^{1/2}}$

    Simplifying, you get:

    $\dfrac{a^2}{(a^2-s^2-t^2)^{1/2}}$

    So, your whole integral simplifies to:

    $\displaystyle \int\int_D \dfrac{a^2}{(a^2-s^2-t^2)^{1/2}}dsdt$

    Does that help?
    Last edited by SlipEternal; March 17th 2014 at 12:40 PM.
    Thanks from gralla55
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  3. #3
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    Re: Surface integral problem

    You are correct! But it's just a typo, the Normal vector should still be the right one.
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  4. #4
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    Re: Surface integral problem

    Quote Originally Posted by gralla55 View Post
    You are correct! But it's just a typo, the Normal vector should still be the right one.
    Very true. I just edited my post above.
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  5. #5
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    Re: Surface integral problem

    And, you don't need to convert to polar coordinates if you don't want. You can use:

    $\displaystyle \iint_D \dfrac{a^2}{\sqrt{a^2-s^2-t^2}}dsdt = a^2 \int_{-a}^a \left( \int_{-\sqrt{a^2-t^2}}^{\sqrt{a^2-t^2}} \dfrac{ds}{\sqrt{(\sqrt{a^2-t^2})^2-s^2}} \right)dt$

    The inner integral is just $\left. \arcsin\left(\dfrac{s}{\sqrt{a^2-t^2}}\right) \right]_{-\sqrt{a^2-t^2}}^{\sqrt{a^2-t^2}} = \arcsin(1)-\arcsin(-1) = \pi$.
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  6. #6
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    Re: Surface integral problem

    Thanks a lot!! That last one I would never have spotted on my own, I guess I still need lots of practice.
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